The following matrix $A_n$ is an $(n-1) \times n$ matrix. I need to solve $A_nX = 0$, where $X = (x_1, \cdots, x_n)$. \begin{equation} A_n = \begin{bmatrix} b_2 & a_2 & b_3 & 0 & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & b_3 & a_3 & b_4 & 0 & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & b_4 & a_4 & b_5 & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & 0 & b_5 & a_5 & b_6 & 0 & \cdots & 0 & 0\\ 0 & 0 & 0 & 0 & b_6 & a_6 & b_7 & \cdots & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & \ddots & \ddots & \ddots & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & \ddots & \ddots & \ddots & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & \ddots & \ddots & b_{n-1} & a_{n-1} & b_n\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots & b_n & a_n\\ \end{bmatrix} \end{equation}
Since $$ -\dfrac{b_n}{a_n}x_{n-1} = x_n, $$ so $x_n$ is a free variable and thus there is always a solution to this system of equations. Let us denote $x_m \sim x_j$ if we can write $x_j$ in terms of $x_m$. Therefore $x_{n-1} \sim x_n$.
By substituting $x_n$ by $x_{n-1}$ in $(n-1)$th row, it implies $x_{n-2} \sim x_{n-1}$ . Therefore we will end up with $x_1 \sim x_2$. My problem is to find the formula, say $\phi$, so that $x_2 = \phi x_1$. Now
I need a hint to propose a method that can help to solve this problem.
Cramer's rule seems doesn't work because the right hand side is zero vector.
Two ingredients
$AX=0$ is a kernel issue. As $A$ is $(n-1) \times n$, assuming that $A$ has maximal rank $n-1$, by rank-nullity theorem, this kernel is 1-dimensional. Therefore, it suffices to find a non-zero vector $V$ of this kernel ; the other elements will be $aV$ (with $a$ a free variable). How to do it ? By...
Block computation. Consider decomposition $A=[U|B]$ where $U$ is the first column and $B$ a $(n-1) \times (n-1)$ matrix. In the same way, decompose your column vector with unknowns $x_i$ into $\binom{x_1}{V}$ where $V=(x_2 \cdots x_n)^T$. In this way, system $AX=0$ becomes :
$$[U|B]\binom{x_1}{V}=0 \iff Ux_1+BV=0 \iff BV=-x_1U$$
which is a $(n-1) \times (n-1)$ system. Taking arbitrarily $x_1=1$ and solving the system (we assume that $B$ is "full-rank") will give a solution $V=V_s$. From this solution, you will get the basis of the kernel under the form :
$$\binom{x_1}{V_s}=\binom{1}{V_s}$$